Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-6x-3y &= 6 \\ 2x-3y &= -6\end{align*}$
Begin by moving the $y$ -term in the second equation to the right side of the equation. $2x = 3y-6$ Divide both sides by $2$ to isolate $x$ $x = {\dfrac{3}{2}y - 3}$ Substitute this expression for $x$ in the first equation. $-6({\dfrac{3}{2}y - 3}) - 3y = 6$ $-9y + 18 - 3y = 6$ Simplify by combining terms, then solve for $y$ $-12y + 18 = 6$ $-12y = -12$ $y = 1$ Substitute $1$ for $y$ in the top equation. $-6x-3( 1) = 6$ $-6x-3 = 6$ $-6x = 9$ $x = -\dfrac{3}{2}$ The solution is $\enspace x = -\dfrac{3}{2}, \enspace y = 1$.